topological vector lattice
A topological vector lattice $V$ over $\mathbb{R}$ is

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a Hausdorff^{} topological vector space^{} over $\mathbb{R}$,

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a vector lattice, and

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locally solid. This means that there is a neighborhood base of $0$ consisting of solid sets.
Proposition 1.
A topological vector lattice $V$ is a topological lattice.
Before proving this, we show the following equivalence on the continuity of various operations^{} on a vector lattice $V$ that is also a topological vector space.
Lemma 1.
Let $V$ be a vector lattice and a topological vector space. The following are equivalent^{}:

1.
$\vee :{V}^{2}\to V$ is continuous^{} (simultaneously in both arguments)

2.
$\wedge :{V}^{2}\to V$ is continuous (simultaneously in both arguments)

3.
${}^{+}:V\to V$ given by ${x}^{+}:=x\vee 0$ is continuous

4.
${}^{}:V\to V$ given by ${x}^{}:=x\vee 0$ is continuous

5.
$\cdot :V\to V$ given by $x:=x\vee x$ is continuous
Proof.
$(1\iff 2)$. If $\vee $ is continuous, then $x\wedge y=x+yx\vee y$ is continuous too, as $+$ and $$ are both continuous under a topological vector space. This proof works in reverse too. $(1\Rightarrow 3)$, $(1\Rightarrow 4)$, and $(3\iff 4)$ are obvious. To see $(4\Rightarrow 5)$, we see that $x={x}^{+}+{x}^{}$, since ${}^{}$ is continuous, ${}^{+}$ is continuous also, so that $\cdot $ is continuous. To see $(5\Rightarrow 4)$, we use the identity^{} $x={x}^{+}{x}^{}$, so that $x=(x+{x}^{})+{x}^{}$, which implies ${x}^{}=\frac{1}{2}(xx)$ is continuous. Finally, $(3\Rightarrow 1)$ is given by $x\vee y=(xy+y)\vee (0+y)=(xy)\vee 0+y={(xy)}^{+}+y$, which is continuous. ∎
In addition, we show an important inequality that is true on any vector lattice:
Lemma 2.
Let $V$ be a vector lattice. Then $\mathrm{}{a}^{\mathrm{+}}\mathrm{}{b}^{\mathrm{+}}\mathrm{}\mathrm{\le}\mathrm{}a\mathrm{}b\mathrm{}$ for any $a\mathrm{,}b\mathrm{\in}V$.
Proof.
${a}^{+}{b}^{+}=({b}^{+}{a}^{+})\vee ({a}^{+}{b}^{+})=(b\vee 0a\vee 0)\vee (a\vee 0b\vee 0)$. Next, $a\vee 0b\vee 0=(b+(a\wedge 0)\vee (a\wedge 0)=((ba)\wedge b)\vee (a\wedge 0)$ so that ${a}^{+}{b}^{+}=((ba)\wedge b)\vee (a\wedge 0)\vee ((ab)\wedge a)\vee (b\wedge 0)\le (ba)\vee (a\wedge 0)\vee (ab)\vee (b\wedge 0)$. Since $(ba)\vee (ab)=ab$ and $a\vee 0$ are both in the positive cone^{} of $V$, so is their sum, so that $0\le (ba)\vee (ab)+(a\vee 0)=(ba)\vee (ab)(a\wedge 0)$, which means that $(a\wedge 0)\le (ba)\vee (ab)$. Similarly, $(b\wedge 0)\le (ba)\vee (ab)$. Combining these two inequalities, we see that ${a}^{+}{b}^{+}\le (ba)\vee (a\wedge 0)\vee (ab)\vee (b\wedge 0)\le (ba)\vee (ab)=ab$. ∎
We are now ready to prove the main assertion.
Proof.
To show that $V$ is a topological lattice, we need to show that the lattice^{} operations meet $\wedge $ and join $\vee $ are continuous, which, by Lemma 1, is equivalent in showing, say, that ${}^{+}$ is continuous. Suppose $N$ is a neighborhood base of 0 consisting of solid sets. We prove that ${}^{+}$ is continuous. This amounts to showing that if $x$ is close to ${x}_{0}$, then ${x}^{+}$ is close to ${x}_{0}^{+}$, which is the same as saying that if $x{x}_{0}$ is in a solid neighborhood $U$ of $0$ ($U\in N$), then so is ${x}^{+}{x}_{0}^{+}$ in $U$. Since $x{x}_{0}\in U$, $x{x}_{0}\in U$. But ${x}^{+}{x}_{0}^{+}\le x{x}_{0}$ by Lemma 2, and $U$ is solid, ${x}^{+}{x}_{0}^{+}\in U$ as well, and therefore ${}^{+}$ is continuous. ∎
As a corollary, we have
Proposition 2.
A topological vector lattice is an ordered topological vector space.
Proof.
All we need to show is that the positive cone is a closed set^{}. But the positive cone is defined as $\{x\mid 0\le x\}=\{x\mid {x}^{}=0\}$, which is closed since ${}^{}$ is continuous, and the positive cone is the inverse image^{} of a singleton, a closed set in $\mathbb{R}$. ∎
Title  topological vector lattice 

Canonical name  TopologicalVectorLattice 
Date of creation  20130322 17:03:51 
Last modified on  20130322 17:03:51 
Owner  CWoo (3771) 
Last modified by  CWoo (3771) 
Numerical id  6 
Author  CWoo (3771) 
Entry type  Definition 
Classification  msc 06F20 
Classification  msc 46A40 
Defines  locally solid 